The ends of a stretched wire of length `L` are fixed at `x = 0 and x = L`. In one experiment, the displacement of the wire is `y_(1) = A sin(pi//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2) = A sin (2pix//L ) sin 2omegat` and energy is `E_(2)`. Then

Options

A. `E_2 = E_1`
B. `E_2 = 2E_1`
C. `E_2 = 4E_1`
D. `E_2 = 16E_1`

Answer

C

Explanation

We know that `E prop A^(2) v^(2)`, where `A` = amplitude and `v` = frequency. Also, `omega = 2 pi v = omega prop v`
In case 1 : Amplitude `= A` and `v_(1) = v`
In case 2 : Amplitude `= A` and `v_(2) = 2v`
`:. (E_(2))/(E_(1)) =(A^(2)v_(2)^(2))/(A^(2)v_(1)^(2)) =4 rArr E_(2)=4E_(1)`.